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within the equation 2x-3*4 you have 2x-12 but if you brake it down more you will have x= -6
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∙ 14y ago
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What is 2x to the third power plus 8x to the second power plus 3x plus 12 when factoring each polynomial by grouping?
2x3 + 8x2 + 3x + 12 = (2x3 + 8x2) + (3x + 12) = 2x2(x + 4) + 3(x + 4) = (x + 4)(2x2 + 3) Since you have asked this question I am assuming that you are not yet at a level where you might want (2x2 + 3) factorised into its imaginary factors.
What is the degree of 4x6 - 2x3 x - 3 x3?
the answer is 6
Quotient and remainder of 3X4 plus 2X3 minus X2 minus x minus 6 divided by X2 plus 1?
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
What is the degree of the polynomial 2x7 plus 4 - 3x3 plus 5x8 - 4x?
I'm going to assume the polynomial in question is 2x7+(3-2x3)+(5x8-4x) Expanding out the polynomial: 2x7+3-2x3+5x8-4x Order the terms by powers of x: 5x8+2x7-2x3-4x+3 Since 8 is the highest power of x, the degree of the polynomial is 8.
Can someone show you the steps in finding the real roots of 2x3-42x plus 40?
Factor out a 2. 2(x3 - 21x + 20) That factors to 2(x - 1)(x - 4)(x + 5) The roots are -5, 1 and 4
X4-2x3 plus 2x2 plus x plus 4 divided by x2 plus x plus 1 solutions?
(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
How do you factor 2x3-42x2-200x?
2x(x + 4)(x - 25)
What are the solutions to the equation 2x 3 plus 4 equals 17?
2x3 + 4 = 17 2x3 = 13 x3 = 13 / 2 x = (13 / 2)1/3 x = 521/3 / 2
What is the value of n in the expression 2 x cubed plus nx squared -x -4 which when divided by x -2 leaves 6 remaining and giving a reason why your answer is correct?
2x3+nx2-x-4 Using the remainder theorem: n = -1 So the expression is now: 2x3-x2-x-4 When divided by x-2 = 2x2+3x+5 remainder 6 or 6 over x-2 The answer is correct because: (x-2)*(2x2+3x+5+6/x-2) = 2x3-x2-x-4
Factor of 2x3 plus 16?
2x3 + 16 = 2(x3 + 8) = 2(x3 + 23) = 2(x + 2)(x2 - 2x + 22) = 2(x + 2)(x2 - 2x + 4)
What is the factor 2x3-8x2 plus 6x?
What are the factors? 2x3 - 8x2 + 6x = 2x(x - 1)(x - 3).
What is 2x3 plus 11x equals 6x?
your equation is this... 2x3 + 11x = 6x 2x3 + 5x = 0 x(2x2 + 5) = 0 x = 0 and (5/2)i and -(5/2)i
2x3 plus 2x2 - 12x?
Do you want its factorisation? 2x3 + 2x2 - 12x = 2x(x2 + x - 6) = 2x(x + 3)(x - 2).
How do you factor a four term polynomial 8xtothe3rdplus12xtothe2nd-44x-24?
8x3 + 12x2 - 44x - 24 = 4(2x3 + 3x2 - 11x - 6) = 4(2x3 - 4x2 + 7x2 - 14x + 3x - 6) = 4[2x2(x - 2) + 7x(x - 2) + 3(x - 2)] = 4(x - 2)(2x2 + 7x + 3) = 4(x - 2)(2x2 + 6x + x + 3) = 4(x - 2)[2x(x + 3) + 1(x + 3)] = 4(x - 2)(x + 3)(2x + 1)
What is the f dashed x or deriative of 1 over 2xcubed plus 5?
f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this
What is the factorization of the trinomial -2x3 plus 2x2 plus 12x?
-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)
What is the factorization of this trinomial -2x3 - 2x2 12x?
I assume your problem is: -2x3-2x2+12x-2x(x2-x-6)-2x(x-3)(x+2)
